What is the solution for 10y^{''}-10y^'+125y=0 ?

The solution for 10y^{''}-10y^'+125y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))

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10y^{''}-10y^'+125y=0

10 y^{^{''}} - 10 y^{^{'}} + 125 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

10 y^{''} (t) - 10 y^{'} (t) + 125 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

x \to \infty lim ((1 + \frac{1}{6 x})^{5 x})

3 t^{2}

\int_{0}^{3} 5 e^{- 3 x} d x

\int \frac{1 - x}{1 - x} d x

x \to + (- \infty) + lim (- x^{2})

What is the solution for 10y^{''}-10y^'+125y=0 ?
The solution for 10y^{''}-10y^'+125y=0 is y=e^{t/2}(c_{1}cos((7t)/2)+c_{2}sin((7t)/2))